## Elementary Algebra

{$-12,2$}
Using the rules for factoring trinomials, we obtain: $y^{2}+10y=24$ $y^{2}+10y-24=0$ $y^{2}-2y+12y-24=0$ $y(y-2)+12(y-2)=0$ $(y-2)(y+12)=0$ $(y-2)=0$ or $(y+12)=0$ $y=2$ or $y=-12$ Therefore, the solution set is {$-12,2$}.